Question

If $\tan {{25}^{\circ }}=x$, prove that $\dfrac{\tan {{155}^{\circ }}-\tan {{115}^{\circ }}}{1+\tan {{155}^{\circ }}\tan {{115}^{\circ }}}=\dfrac{1-{{x}^{2}}}{2x}$.

Answer 1

Hint: To prove the given equation, we are going to write the L.H.S in the expression in x and then prove that both the sides are equal. This conversion is going to be achieved by writing $\tan {{155}^{\circ }}=\tan \left( {{180}^{\circ }}-{{25}^{\circ }} \right)$ and writing $\tan {{115}^{\circ }}=\tan \left( {{90}^{\circ }}+{{25}^{\circ }} \right)$ in the L.H.S of the given equation. There are trigonometric conversions which states that: $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $ and $\tan \left( {{90}^{\circ }}+\theta \right)=-\cot \theta $. These conversions we are going to use the L.H.S and hence, will prove the given equation.

Complete step by step answer:
The equation given in the above problem which we are asked to prove is as follows:
$\dfrac{\tan {{155}^{\circ }}-\tan {{115}^{\circ }}}{1+\tan {{155}^{\circ }}\tan {{115}^{\circ }}}=\dfrac{1-{{x}^{2}}}{2x}$
We have also given that: $\tan {{25}^{\circ }}=x$ so using this relation we are going to make the L.H.S equal to R.H.S.
We can write $\tan {{155}^{\circ }}=\tan \left( {{180}^{\circ }}-{{25}^{\circ }} \right)$ and $\tan {{115}^{\circ }}=\tan \left( {{90}^{\circ }}+{{25}^{\circ }} \right)$ in the L.H.S of the given equation and we get,
$\begin{align}
  & \dfrac{\tan {{155}^{\circ }}-\tan {{115}^{\circ }}}{1+\tan {{155}^{\circ }}\tan {{115}^{\circ }}}=\dfrac{1-{{x}^{2}}}{2x} \\
 & \Rightarrow \dfrac{\tan \left( {{180}^{\circ }}-{{25}^{\circ }} \right)-\tan \left( {{90}^{\circ }}+{{25}^{\circ }} \right)}{1+\tan \left( {{180}^{\circ }}-{{25}^{\circ }} \right)\tan \left( {{90}^{\circ }}+{{25}^{\circ }} \right)}=\dfrac{1-{{x}^{2}}}{2x} \\
\end{align}$
We know the trigonometric conversions which states that:
$\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $ and $\tan \left( {{90}^{\circ }}+\theta \right)=-\cot \theta $
Substituting $\theta ={{25}^{\circ }}$ in the above equations we get,
$\tan \left( {{180}^{\circ }}-{{25}^{\circ }} \right)=-\tan {{25}^{\circ }}$ and $\tan \left( {{90}^{\circ }}+{{25}^{\circ }} \right)=-\cot {{25}^{\circ }}$
Using the above equations in the given equation we get,
$\dfrac{-\tan \left( {{25}^{\circ }} \right)+\cot \left( {{25}^{\circ }} \right)}{1+\tan \left( {{25}^{\circ }} \right)\cot \left( {{25}^{\circ }} \right)}=\dfrac{1-{{x}^{2}}}{2x}$
Now, we know that $\cot \theta =\dfrac{1}{\tan \theta }$ using this relation in the above equation we get,
$\dfrac{-\tan \left( {{25}^{\circ }} \right)+\dfrac{1}{\tan {{25}^{\circ }}}}{1+\tan \left( {{25}^{\circ }} \right)\left( \dfrac{1}{\tan {{25}^{\circ }}} \right)}=\dfrac{1-{{x}^{2}}}{2x}$
In the denominator of the L.H.S of the above equation, $\tan {{25}^{\circ }}$ will get cancelled out from the numerator and the denominator and we get,
$\dfrac{-\tan \left( {{25}^{\circ }} \right)+\dfrac{1}{\tan {{25}^{\circ }}}}{1+1}=\dfrac{1-{{x}^{2}}}{2x}$
Substituting $\tan {{25}^{\circ }}=x$ in the L.H.S of the above equation we get,
$\dfrac{-x+\dfrac{1}{x}}{2}=\dfrac{1-{{x}^{2}}}{2x}$
Now, taking $x$ as L.C.M in the L.H.S of the above equation we get,
$\begin{align}
  & \dfrac{-x\left( x \right)+1}{2x}=\dfrac{1-{{x}^{2}}}{2x} \\
 & \Rightarrow \dfrac{-{{x}^{2}}+1}{2x}=\dfrac{1-{{x}^{2}}}{2x} \\
\end{align}$
Rearranging L.H.S of the above equation we get,
$\dfrac{1-{{x}^{2}}}{2x}=\dfrac{1-{{x}^{2}}}{2x}$
As you can see that L.H.S is equal to R.H.S so we have verified the given equation.

Note: The mistake that could be possible in the above problem is that while writing the trigonometric conversions, you might forget to write the negative sign after the conversions.
$\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $ and $\tan \left( {{90}^{\circ }}+\theta \right)=-\cot \theta $
There is a negative sign you can see in the above conversions so there is a trick and on using that you will not forget the negative sign and the trick is as you can see that the angles written inside $\tan $ lies in the second quadrant and in the second quadrant $\tan $ is negative.